Secondly, even if the perpendicular FD may not meet the vertex D, the case may be proved by propositions 15 and 13. Similarly, make the rectangle LMNP equal to the triangle ADB. And the two rectangles LMNP and DFCE are together equal to the figure ABCD. NP Of the two rectangles DFCE and LMNP place one of the sides NP of the one on DE of the other, so that, CE and PM may be in the same straight line. Produce LN to the point O in the straight line FC. Join OE and produce OE. Produce LM to Q so that OQ and LQ may meet. From Q draw QR parallel to MC. Produce OC to R and NP to S. Because LR is a parallelogram (Def. 24), therefore the complements NM and ER are equal (Prop. 20). Therefore the parallelograms DC and CS are together equal to the rectilineal figure ABCD. But DC and CS are together equal to the retangle DR (Ax. 8), wherefore DR is described a rectangle equal to the rectilineal figure ABCD Q. E. F. Cor. 1. A parallelogram on half the base of a triangle and between the same parallels is equal to the triangle. Cor. 2. Given two sides, to make a rectangle. Place the two sides so, that they may be contiguous and perpendicular to each other. Complete the figure with two other perpendiculars drawn from the other extremities of the sides. The rectangle is said to be contained by its two contiguous sides. Cor. 3. Given a side, to make a square. If the perpendiculars drawn from the extremities of the side are equal to the side and the other extremities of the perpendicular be joined, the rectangle thus formed is a square. The squar of a straight line AB is usually expressed by saying “square AB, or the rectangle AB. AB." PROP. XXII. THEOREM. (E. 6. 16).. If four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle oontained by the means ; and conversely, if two rectangles are equal, the straight lines containing them are proportionals. Let the straight lines AB, CD, CF and AC be four proportionals, i.e., AB t CD as CF to AC. The rectangle contained by AB and AC is equal to the rectangle contained by CD and CF. Make the rectangles AB. AC and CD. CF (21. cor. 2). Because AB to CD as CF to AC and because the angles CAB and FCD are equal (Ax. 11), therefore the rectangles AC. AB and CF. CD are equal (P. 19). And conversely, because (hyp.) the rectangles AC. AB and CF.CD are equal, and have their angles CAB and FCD also equal (Ax. 11) therefore AB to CD as CF to AC (P. 19). Wherefore if four straight lines &c. Q.E.D. Cor. It is manifest that if CDCF then the square of CD is equal to the rectangle contained by AB and AC. Therefore if three straight lines be proportionals, the rectangle contained by the extremes is equal to the square of the mean. (E. 6. 17), and conversely, if the rectangle contained by the extremes is equal to the square of the mean, the three straight lines are proportionals. PROP. XXIII. THEOREM. (E. 1. 32). If a side of a triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are together equal to two right angles. Let ABC be a triangle, and let one of its sides BC be produced to D. Then the exterior angle ACD is equal B A to the two interior and opposite angles CAB, ABC. and the three interior angles ABC, CAB and ACB are equal to two right angles. Through the point C, draw the straight line CE parallel to the side BA (P. 9). Because CE is parallel to BA, and AC meets them, the angle ACE is equal to the alternate angle BAC (P. 7). Again, because CE is parallel to AB and BD falls upon them, the exterior angle ECD is equal (P. 8) to the interior and opposite angle ABC. But the angle ACE was shown to be equal to the angle BAC. Therefore the whole exterior angle ACD is equal to the two interior and and opposite angles CAB and ABC. To each of these equals, add the angle ACB. Therefore the two angles ACD, ACB are equal to the three angles CAB, ABC, ACB. But the two angles ACD, ACB are equal to two right angles (5. cor. 2). Therefore also the three angles CAB, ABC and ACB are equal to two right angles. Wherefore if a side of any triangle &c. Q. E. D. Cor. 1. All the interior angles of any rectilineal figure together with four right angles are equal to twice as many right angles as the figure has sides, D Let ABCDE be any rectilineal figure. All the interior angles ABC, BCD, &c. together with four right angles are equal to twice as many right angles as the figure has sides. Divide the rectilineal figure ABCDE into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. A F Because the three interior angles are equal (P. 23) to two right angles, and there are as many triangles in the figure as it has sides, all the angles of these triangles are equal to twice as many right angles as the figure has sides. But all the angles of these triangles are equal to the interior angles of the figure together with the angles at the point F, which can be shown to be equal to four right angles. Therefore all the angles of these triangles are equal (Ax. 1) to the interior angles of the figure together with four right angles. But it has been proved that all the angles of these triangles are equal to twice as many right angles as the figure has sides. Therefore all the angles of the figure together with four right angles are equal to twice as many right angles as the figure has sides. Cor. 2. All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles. Because the interior angle ABC, and its adjacent exterior angle ABD, are (5. Cor. 2) together equal to two right angles. Therefore all the interior angles, together with all the exterior angles of the figure, are equal to A B twice as many right angles as the figure has sides. But it has been proved by the foregoing corollary, that all the interior angles together with four right angles are equal to twice as many right angles as the figure has sides. Therefore all the interior angles together with all the exterior angles are equal (Ax. 1) to all the interior angles and four right angles. Take from these equals all the interior angles. Therefore all the exterior angles of the figuré are equal to four right angles. PROP. XXIV. THEOREM. If a straight line be bisected and from the point of section, a straight line be drawn equal to one of the parts, the angle made at the vertex of the straight line so drawn, by the sides drawn from it to the extremities of the base, is a right angle; and conversely, in a right-angled triangle if a straight line be drawn from the right angle bisecting the base, the straight line is equal to half the base. First, Let AB be a straight line bisected at C, and CD be drawn from C equal to AC or CB. Draw DA, DB. The angle ADB is a right angle. Because AC is equal to CD (hyp), the angle DAC is equal to the angle ADC (P. 4). For the same reason the angle CDB is equal to the angle CBD. Therefore the whole angle ADB is equal to the two angles DAB and DBA. But angles ADB, DAB and DBA are equal to two right angles (P. 23). And because the angle ADB is half of the sum, i.e., two right angles, therefore is equal to one right angle. |